A particle of mass 1 kg is subjected to a force which depends on the position as F → = - k x i ^ + y j ^ kg m s - 2 with k = 1 kg s - 2 . At time t = 0 , the particle's position r → = 1 2 i ^ + 2 j ^ m and its velocity v → = - 2 i ^ + 2 j ^ + 2 π k ^ m s - 1 . Let v x and v y denote the x and the y components of the particle's velocity, respectively. Ignore gravity. When z = 0 . 5 m , t

Step-by-step solution

Given here: F → = - k x i ^ + y j ^   kg   m   s - 2 and m = 1   kg In x-direction, F x = - x = m a x So, acceleration, a x = d 2 x d t 2 = - x Now, for particle executing SHM, displacement along x-direction is ⇒ x = A x sin ω t + ϕ x , here, angular frequency, ω = 1   rad   s - 1 and velocity, v x = A x ω cos ω t + ϕ x Given at t = 0 ,   x = 1 2   m and v x = - 2   m   s - 1 So, putting the values, we get 1 2 = A x sin &#98